356

STOP

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66

(b) Let (-9-4i/3+4i/2+2i/6)» = V(-m+n).
v[(-m+n)(-m-n)J=Vi(-m) i -n i 1 = V [73+40 /3]=»/[p+?].
i/[(p+?)(p-ff)]=28.

Let i/(p+g) = i/ (a!+28)+i/aj. Then 2x+23=73, or £=25.
•••l/(p + g)=4|/3+5.

Let v 7 (w-ro) =-1/ (as +4i/3 f 5)+v / «. Then2a>+4i/3+5--9-4i/3,
or *=— 7— 4i/3).

.-. |/(w-m)=i/(-7-4|/3)-]/-2.

i/[(-7-4|/3)(-7+4 1 /3)]=l.

Let |/(-7-4i/3)=i/(a;+l)+i/x; then 2b+1=-7, »=-4.

.••i/(-7-4i/3)=i/(-3)+2|/(-l).

.M/(«-w)=v/(-3)+2|/(-l)-,/(-2).

Also solved by G. I. Hopkins and A. H. Holmes.
332. Proposed by C. N. SCHMALL, New York City.

Solve the quadratic, x 2 +ax+b=0, without completing the square.

Solution by AKTEMAS MARTIN, LL. D., Washington, D. C.

Assume y— ha=x, and substitute in the given quadratic and it becomes

(y-ha) z +a(y-ia)+b^0, or y*-ia 2 +b=0;

whence y= ± v {la 2 —b) and x = ± i/Ja 2 — 6) — Ja.

See Mathematical Magazine, Vol. I, No. 9 (January, 1884) , p. 146.

Solved similarly by V. M. Spunar , Levi S. Shively and the Proposer.

Professor Hopkins, in his solution, made use of the principle that the sum of the roots is equal to the coeffi-
cient of x with sign changed, and the product of the roots is equal to the final term.
S. Lef sehetz sent in solutions of 327 and 328 too late for credit in last issue.

GEOMETRY.

356. Proposed by G. I. HOPKINS, Manchester, N. H.

Required to construct a triangle having given, base, vertical angle, and difference of
other two sides.

I. Solution by J. M. ARNOLD, Crompton, R. I.

Let x=the longer side, then x— d=the shorter side. Let A=the ver-
tical angle. Then

x 2 - {xcosA) s = (x-d) i - (6-a;cos^l) 2
which gives

(2bcosA-2d)x=b 2 -d 2 .

Dividing by 4 and putting in the form of a proportion

67

(ibcosA—hd)

b-d^ b ±d
2 2

x.

Construction. Draw AC equal to b, draw the indefinite line AW,
making an angle at A equal to the given angle. On A W lay off AD equal
to hb, and draw DE perpendicular to AC. From E lay off EI equal to — Id.
Then will AI equal IbcosA-hd.

On AW lay off 4ff equal to

b + d

d

and draw HI. On AC lay off AM

equal to
Join £C.

and through M draw a line parallel to HI meeting AW at B.

Then will ABC be the triangle required.

Similarly solved by G. B. M. Zerr.

II. Solution by J. SCHEFFER, A. M., Hagerstown, Md.

On the given base, BC, as a chord, describe a circle 0, containing the
segment whose angle contains the angle=90°
~\~hA, A being the given vertical angle, and also a
circle 0', the segment of which contains the angle
A. Make BD= given difference of sides; extend
BD to A, where it cuts the circumference of circle
0'. Draw AC; then ABC is the required triangle.

For, AADC=90°-hA.

.: AACD=90°-hA; :.AD=AC; :-AB-AC
=BD~ given difference, which proves construction.

Solved similarly C. N. Schmall and H. C. Feemster.
357. Proposed by E. R. HOYT, St. Louis, Mo.

A room is 30 feet long, 12 feet wide, and 12 feet high. At one end of the room, 3
feet from the floor, and midway from the sides, is a spider. At the other end, 9 feet from
the floor, and midway from the sides, is a fly. Determine the shortest path by way of the
floor, ends, sides, and ceiling, the spider can take to capture the fly.

Solution by G. B. M. ZERR, A. M., Ph. D., Philadelphia, Pa.

Suppose the six sides of the room spread out in one plane as in the
figure, the floor being the second rectangle from the bottom, and let outdis-
tance of spider from floor, 12 —x the distance of the fly, x<6. There are
three courses for the spider to take.
First, the route SC=30+x+12-a;=42 feet... (1).

Second, the route SB=\/[(Sb) 2 + (Bb)*]= ] /[(26+x)* + (18-x) 2 ')...(2).
Third, the route SA=i/[(Sa) s + Ua) 2 ] = |/[(30+2a;) s + (24) 2 ]...(3).

Let (30+2a0 2 +576 = (36+a0 2 + (18-a0 3 .